My Math Forum  

Go Back   My Math Forum > High School Math Forum > Pre-Calculus

Pre-Calculus Pre-Calculus Math Forum


Thanks Tree4Thanks
  • 2 Post By v8archie
  • 1 Post By skipjack
  • 1 Post By SDK
Reply
 
LinkBack Thread Tools Display Modes
August 15th, 2018, 07:32 PM   #1
Senior Member
 
happy21's Avatar
 
Joined: Jan 2012

Posts: 123
Thanks: 2

Evaluating a limit

Hi,

Please evaluate the following limit:

$\displaystyle \lim_{x->0}\left ( \tfrac{1}{x^{2}}-\cot^{2}x \right )$

Thanks.

Last edited by skipjack; August 15th, 2018 at 08:41 PM.
happy21 is offline  
 
August 15th, 2018, 08:03 PM   #2
SDK
Senior Member
 
Joined: Sep 2016
From: USA

Posts: 520
Thanks: 293

Math Focus: Dynamical systems, analytic function theory, numerics
The limit is 2/3.
SDK is offline  
August 15th, 2018, 08:27 PM   #3
Senior Member
 
happy21's Avatar
 
Joined: Jan 2012

Posts: 123
Thanks: 2

Quote:
Originally Posted by SDK View Post
The limit is 2/3.
Can you please explain?

Last edited by skipjack; August 15th, 2018 at 09:29 PM.
happy21 is offline  
August 15th, 2018, 08:44 PM   #4
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,509
Thanks: 2514

Math Focus: Mainly analysis and algebra
Assuming that the limit exists and that therefore $$\lim_{x \to 0} \left(\frac1{x^2}-\cot^2{x}\right) = L$$
setting $2t=x$ we have that
\begin{align}L &= \lim_{t \to 0} \left(\frac1{(2t)^2}-\cot^2{2t}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{\sin^2{2t}}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{(2\sin{t}\cos{t})^2}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{(\cos^2{t}-\sin^2{t})^2}{4\sin^2{t}\cos^2{t}}\right) \\
4L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
3L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) - \lim_{t \to 0} \left(\frac1{t^2}-\cot^2{t}\right) \\
&=\lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} - \frac1{t^2}+\cot^2{t}\right) \\
&=\lim_{t \to 0} \left(\frac{\cos^2{t}}{\sin^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(\frac{\cos^4{t}}{\sin^2{t}\cos^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(\frac{2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{ t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(2+\tan^2{t} \right) \\
&=2 \\
L&=\frac23
\end{align}

Thanks to this for inspiration.
Thanks from happy21 and topsquark
v8archie is offline  
August 15th, 2018, 09:21 PM   #5
Senior Member
 
happy21's Avatar
 
Joined: Jan 2012

Posts: 123
Thanks: 2

Quote:
Originally Posted by v8archie View Post
Assuming that the limit exists and that therefore $$\lim_{x \to 0} \left(\frac1{x^2}-\cot^2{x}\right) = L$$
setting $2t=x$ we have that
\begin{align}L &= \lim_{t \to 0} \left(\frac1{(2t)^2}-\cot^2{2t}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{\sin^2{2t}}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{(2\sin{t}\cos{t})^2}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{(\cos^2{t}-\sin^2{t})^2}{4\sin^2{t}\cos^2{t}}\right) \\
4L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
3L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) - \lim_{t \to 0} \left(\frac1{t^2}-\cot^2{t}\right) \\
&=\lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} - \frac1{t^2}+\cot^2{t}\right) \\
&=\lim_{t \to 0} \left(\frac{\cos^2{t}}{\sin^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(\frac{\cos^4{t}}{\sin^2{t}\cos^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(\frac{2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{ t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(2+\tan^2{t} \right) \\
&=2 \\
L&=\frac23
\end{align}

Thanks to this for inspiration.
Brilliant! This method is very nice though uncommon, I think!

Thx.
happy21 is offline  
August 15th, 2018, 10:06 PM   #6
Global Moderator
 
Joined: Dec 2006

Posts: 19,957
Thanks: 1844

$\displaystyle \begin{align*}\cot^2\!x &= 1/(\sin^2\!x) - 1 \\
&= 1/(x^2(1 - x^2/3 \,+\; ...)) - 1 \\
&= (1/x^2)(1 + x^2/3 \,+\; ...) - 1 \\
&= 1/x^2 - 2/3 \,+\; ...\end{align*}$
so the required limit is 2/3.
Thanks from happy21
skipjack is offline  
August 16th, 2018, 06:06 AM   #7
SDK
Senior Member
 
Joined: Sep 2016
From: USA

Posts: 520
Thanks: 293

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by happy21 View Post
Can you please explain?
If you show me some work I'll show you some work.
Thanks from topsquark
SDK is offline  
August 16th, 2018, 10:59 AM   #8
Senior Member
 
happy21's Avatar
 
Joined: Jan 2012

Posts: 123
Thanks: 2

Quote:
Originally Posted by SDK View Post
If you show me some work I'll show you some work.
Yeah I did from the methods depicted above.
happy21 is offline  
Reply

  My Math Forum > High School Math Forum > Pre-Calculus

Tags
evaluating, limit



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
evaluating this limit mike1127 Calculus 6 March 22nd, 2016 07:14 PM
Evaluating the Limit rasa97 Calculus 5 August 28th, 2015 04:42 PM
There is a shorthand for evaluating this limit ? szz Calculus 1 December 7th, 2014 05:17 PM
Evaluating a limit Aurica Calculus 2 June 21st, 2009 07:34 PM
evaluating limit lefucer Calculus 2 June 28th, 2008 06:11 PM





Copyright © 2018 My Math Forum. All rights reserved.